While working with the following problem in complex analysis:
I have $\varphi(z)=z+\sum_{n=2}^{\infty} a_{n}z^n$ by considering $z_0=0$, i.e. a power series expansion centered around zero. Now if I raise $\varphi^k(z)$ I get $z^k + O(z^{n+1})$, but I should get a linear term in z. What am I missing? For example, for k=2, I have $\varphi^2(z)=z^2+a_{n}z^{n+1}+....$.
I havve an auxillary question in this problem, why is $z_{0} = 0$?
Answer assuming $\Omega$ is simply connected
To your auxiliary question, there’s no reason $z_0=0$, but it’s a smart idea to transform it into zero. Since $\Omega$ is bounded and open$^1$, the Riemann mapping theorem lets you find a conformal map $f:\mathbb D\to\Omega$. This map does not necessarily take $0$ to $x_0$, but by composing with a suitable automorphism of $\mathbb D$, this can be arranged. Define $g=f^{-1}\circ\phi\circ f:\mathbb D\to\mathbb D$ with $$g(0)=f^{-1}(\phi(f(0)))=f^{-1}(\phi(x_0))=f^{-1}(x_0)=0.$$ Thus $g$ satisfies the hypotheses of the Schwarz lemma. Moreover, iterating the chain rule twice, $$g’(0)=(f^{-1})’\circ\phi\circ f(0)\phi’\circ f(0)f’(0)=\frac{\phi’\circ f(0)f’(0)}{f’\circ f^{-1}\circ\phi\circ f(0)}=\frac{\phi’(x_0)f’(0)}{f’\circ f^{-1}\circ\phi(x_0)}=\cdots=1$$ where I’ve skipped some of the easy plugging in of numbers. The Schwarz Lemma tells us that $g$ is in the form $g(z)=\alpha z$ and in particular $\alpha=1$. Thus, returning to the definition of $g$, $\phi(z)=f\circ g\circ f^{-1}(z)=f\circ f^{-1}(z)=z$.
Answer for general (connected) $\Omega$
As suggested by Martin R, we use iterates of $\phi$ to solve the problem in general. By translating the problem, we may assume $z_0=0$ and we can write $$\phi(z)=z+az^m+O(z^{m+1}),$$ where the 0th order term vanishes because $\phi(0)=0$ by assumption. Here $z^m$ is the next nonzero term in the Taylor series, but of course it could be that there are none and $a=0$. We define the iterates of $\phi$ as the composition $\phi^k=\underbrace{\phi\circ\cdots\circ\phi}_k$. Moreover, since $\Omega$ is bounded, $\phi:\Omega\to\Omega$, and consequently $\phi^k:\Omega\to\Omega$, we see that $\phi^k$ must be bounded. Uniform boundedness of a holomorphic family implies it is a normal family by Montel’s theorem, so we conclude $\{\phi^k:k\geq0\}$ is normal.
Next we compute $\phi^k$. From the Taylor approximation above, you can see that $\phi=z+O(z^m)=O(z)$. Therefore $$\phi^2(z)=\phi\circ\phi(z)=(z+az^m+O(z^{m+1}))+a(z+O(z^m))^m+O(z^{m+1}).$$ Taking advantage of the $O(z^{m+1})$ error and using the binomial theorem, this implies $$\phi^2(z)=z+2az^m+O(z^{m+1}).$$ Repeating this, one finds that $$\phi^k(z)=z+kaz^m+O_k(z^{m+1}).$$ So we compute the $m$th derivative$^1$ of the $k$th iterate, $$(\phi^k)^{(m)}(0)=m!ka.$$ If $a\neq0$, these $m$th derivatives at $0$ are not bounded in the family, so no sequence within it can be converging to an analytic function, contradicting the normalness we showed. Therefore $a=0$ and so $\phi=z$ in a neighborhood of $0$, and therefore in the connected component of $\Omega$ containing $0$. (If $\Omega$ is disconnected, there is nothing stopping $\phi$ from having some other behavior on the other components.)
$^1$ Please note that by $O(z^m)$ we mean an expression of the form $z^mg(z)$ where $g(z)$ is analytic. This is stronger than the more common meaning of the notation but we need it to differentiate here.